Question: Solve for $x$ : $ 5|x - 1| - 3 = -4|x - 1| + 2 $
Explanation: Add $ {4|x - 1|} $ to both sides: $ \begin{eqnarray} 5|x - 1| - 3 &=& -4|x - 1| + 2 \\ \\ { + 4|x - 1|} && { + 4|x - 1|} \\ \\ 9|x - 1| - 3 &=& 2 \end{eqnarray} $ Add ${3}$ to both sides: $ \begin{eqnarray} 9|x - 1| - 3 &=& 2 \\ \\ { + 3} &=& { + 3} \\ \\ 9|x - 1| &=& 5 \end{eqnarray} $ Divide both sides by ${9}$ $ \dfrac{9|x - 1|} {{9}} = \dfrac{5} {{9}} $ Simplify: $ |x - 1| = \dfrac{5}{9}$ Because the absolute value of an expression is its distance from zero, it has two solutions, one negative and one positive: $ x - 1 = -\dfrac{5}{9} $ or $ x - 1 = \dfrac{5}{9} $ Solve for the solution where $x - 1$ is negative: $ x - 1 = -\dfrac{5}{9} $ Add ${1}$ to both sides: $ \begin{eqnarray} x - 1 &=& -\dfrac{5}{9} \\ \\ {+ 1} && {+ 1} \\ \\ x &=& -\dfrac{5}{9} + 1 \end{eqnarray} $ Change the ${ + 1}$ to an equivalent fraction with a denominator of $9$ $ x = - \dfrac{5}{9} {+ \dfrac{9}{9}} $ $ x = \dfrac{4}{9} $ Then calculate the solution where $x - 1$ is positive: $ x - 1 = \dfrac{5}{9} $ Add ${1}$ to both sides: $ \begin{eqnarray} x - 1 &=& \dfrac{5}{9} \\ \\ {+ 1} && {+ 1} \\ \\ x &=& \dfrac{5}{9} + 1 \end{eqnarray} $ Change the ${ + 1}$ to an equivalent fraction with a denominator of $9$ $ x = \dfrac{5}{9} {+ \dfrac{9}{9}} $ $ x = \dfrac{14}{9} $ Thus, the correct answer is $x = \dfrac{4}{9} $ or $x = \dfrac{14}{9} $.